Poles of abelian differentials

Question

Let $X$ be a smooth projective curve of genus $g$ over an algebraically closed field $k$. As a corollary of the Riemann-Roch theorem we know that for every abelian differential $\omega$ on $X$ we have $$ \deg(\omega) = 2g - 2. $$

Now assume that $g\geq 2$, then $\deg(\omega)\geq 0$. Does this imply that an arbitrary differential has no poles on the whole $X$? I guess no: a priori this just means that the number of zeros (with multiplicity) is greater than the number of poles. But anyway, is it always possible to find a function $f\in k(X)$ such that the form $f \omega$ has no poles?

Any idea or clarification is welcome!

Answer 1

Yes, you can always find such a rational function $f\in k(X)$. Here is why:

Since the cotangent bundle $\Omega_X$ satisfies $\operatorname {dim} H^0(X,\Omega_X)=g\geq 2$, there certainly exists a regular differential form $ \eta\neq 0\in H^0(X,\Omega_X)$.
As is the case for all line bundles, the quotient $\frac \eta\omega$ of two rational sections (with $ \omega\neq0$) of $\Omega_X$ is a rational function $f\in k(X)$ and we have found our desired functoion $f$ with $\eta=f\omega \in H^0(X,\Omega_X)$, a differential form with no poles.
Notice that the result is true also for $g=1$ but of course false for $g=0$.

[As to your first question, you can of course force any differential form to acquire a pole at any designated point by multiplying it by a rational function having a sufficiently bad pole at that point]

Answer 2

Since $g=p_g=\dim_k \Gamma(X,\omega_X)\geq 2$, we may fix any nonzero regular abelian differential $\omega_0$. Therefore, any rational abelian differential $\omega$ is a rational section of the invertible sheaf $\omega_X$, hence $\omega$ can be represented as $\omega=f\omega_0$ for some $f\in k(X)$, so $\omega/f$ has no poles.

However, $div(\omega)=div(f)+div(\omega_0)$ is not necessarily effective (e.g. $f$ may have poles outside the zeroes of $\omega_0$), so $\omega=f\omega_0$ is not regular in general.